What is the tangent of f(x)=3sqrt(x),(1,3) ?

The tangent of f(x)=3sqrt(x),(1,3) is y= 3/2 x+3/2

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tangent of f(x)=3sqrt(x),(1,3)

tangent of

f (x) = 3 x, (1, 3)

y = \frac{3}{2} x + \frac{3}{2}

Solution steps

Find the slope of

f (x) = 3 x : \frac{df ( x )}{d x} = \frac{3}{2 x}

Compute the slope of

f (x) = 3 x

(1, 3) : m = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(1, 3) : y = \frac{3}{2} x + \frac{3}{2}

y = \frac{3}{2} x + \frac{3}{2}

\int_{- 2}^{4} ∣ - 2 x + 4 ∣ d x

y^{^{'}} - \frac{2}{x} y = 0

\int 6 y e^{2 x y} + x d x

\frac{\partial}{\partial x} (\frac{7 x ^{2}}{( x - y ) ^{2}})

\int_{1}^{t} \frac{3}{x ^{3}} d x

What is the tangent of f(x)=3sqrt(x),(1,3) ?
The tangent of f(x)=3sqrt(x),(1,3) is y= 3/2 x+3/2